Continuity

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close as you want

A function being continuous at a domain number is the opposite of being discontinuous.

The basic idea is that function values are close when the domain numbers are close.

Continuous

The function $f$ is continuous at $a$ if

$a \in Dom_f$, $a$ is in the domain of $f$, which means $f(a) \in \mathbb {R}$, i.e. $f(a)$ exists.
For every $\epsilon > 0$ defining an open interval around $f(a)$, namely $I = (f(a) - \epsilon , f(a) + \epsilon )$, there exists a corresponding $\delta > 0$, such that the image of the open interval $(a - \delta , a + \delta )$ is a subset of $(f(a)-\epsilon , f(a)+\epsilon )$.

For EVERY open interval $I$ around $f(a)$, there exists a corresponding open interval around $a$, such that ALL of the function values are inside $I$.

$\blacktriangleright $ Discontinuity: You can find a single interval around $f(a)$, such that EVERY interval around $a$ contains a domain number whose function value is outside that interval.

$\blacktriangleright $ Continuity: For any and every selected interval around $f(a)$, you can find a corresponding interval around $a$, such that all of their function values are inside the selected interval.

It is going to take some experience to organize our thoughts around every, all, and there exists.

\[ \text {Let } \, f(x) = \begin{cases} 2x-4 & \text { on } (-2, 5) \\ -x + 9 & \text { on } [5, 9) \end{cases} \]

Graph of $y=f(x)$

$f$ is continuous at $3$.

explanation

First, $f(3) = 2$.

Pick ANY open interval around $2$: $(2 - \epsilon , 2 + \epsilon )$, where $\epsilon > 0 $ is small.

Then there is a corresponding interval for the domain. Namely choose $\delta = \frac {\epsilon }{4}$.

Let $I = (3 - \delta , 3 + \delta ) = \left (3 - \frac {\epsilon }{4}, 3 + \frac {\epsilon }{4}\right )$

The image of $I$, $f(I) = f\left (\,\left (3 - \frac {\epsilon }{4}, 3 + \frac {\epsilon }{4}\right )\,\right ) = \left (2 - \frac {\epsilon }{2}, 2 + \frac {\epsilon }{2}\right )$

And, $\left (2 - \frac {\epsilon }{2}, 2 + \frac {\epsilon }{2}\right ) \subset (2 - \epsilon , 2 + \epsilon ) $

ALL of the function values land inside the original $(2 - \epsilon , 2 + \epsilon )$ interval.

This can be done FOR ANY $\epsilon > 0$.

For ANY open interval around $f(a)$, no matter how small, a corresponding open interval around $a$ can be found, such that the function value from that interval around $a$ never jump out of the interval around $f(a)$.

Continuity: No matter what $\epsilon $ is choosen for $f(a)$, you can ALWAYS find a corresponding $\delta $ for $a$.

Discontinuity: There is an $\epsilon $ for $f(a)$, such that there is no $\delta $ for $a$.

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more examples can be found by following this link
More Examples of Continuity

2026-05-31 15:51:41